2 Answers
2

The function $g(x)$ gives no problem, except at $x=-1$. In that sense, it is very different from $f(x)$, which is only defined for $|x|<1$, as explained by Arturo Magidin and Michael Hardy. The calculation you made at the beginning of the post is a bit informal, but is essentially correct for $|x|<1$. (Please see the comment at the end.) And your observation that $g(x)$ cannot always be equal to $x$ is well-founded. You correctly saw that $g(x)$ exists for any positive $x$. Though you did not give a proof, with some work your observation about the fraction being less than $1$ can be turned into a proof. We now solve the problem in detail.

So we can find an explicit formula for $g_n(x)$, where
$$g_n(x)=\frac{\sum_{i=1}^{n}x^i}{1+\sum_{i=1}^{n}x^i}.$$
The result (except when $x=\pm 1$) is
$$g_n(x)=\frac{x(1-x^n)}{1-x^{n+1}}.$$
If $|x|<1$, then $\displaystyle\lim_{n\to\infty} x^n=0$, and therefore $\displaystyle\lim_{x\to\infty} g_n(x)=x$. If $|x|>1$, the limit is $1$.

The case $x=-1$ is hopeless, since the denominator is $0$ for all odd $n$. In the case $x=1$, a separate calculation shows that the limit is $1$.

Comment: The case $|x|>1$ is obvious without calculation. If $|x|>1$, and $n$ is large, then the numerator and denominator of $g_n(x)$ each has very large absolute value. But they differ by $1$, so their ratio is nearly $1$. The case $|x|<1$ is also obvious without much calculation. The numerator of $g_n(x)$ is $x(1+x+\cdots +x^{n-1})$, and the denominator is $1+x+\cdots +x^n$. As long as we know that $\displaystyle\lim_{m\to\infty}(1+x+\cdots +x^m)$ exists, and is non-zero, and that $g_n(x)$ is always defined, we can conclude that $g_n(x)$ has limit $x$. We do not need to know a formula for $\displaystyle\lim_{m\to\infty}(1+x+\cdots +x^m)$.

If $\sum^{\infty}b_i=0$, and $a_i=2b_i$, then both limits in your second expression exist, but the second expression is undefined, whereas the first expression could be 2.
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Gerry MyersonNov 5 '11 at 8:34

Yes, I should have added "...and the denominator is not zero".
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Michael HardyNov 5 '11 at 16:32